Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <stdio.h>
#include <iostream>
#include <string>
#include <sstream>
#include <stack>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#include <cstdlib>
#include <bitset>
#include <tuple>
#include <assert.h>
#include <deque>
#include <bitset>
#include <iomanip>
#include <limits>
#include <chrono>
#include <random>
#include <array>
#include <unordered_map>
#include <functional>
#include <complex>
#include <numeric>
#include <cctype>


template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
//constexpr long long MAX = 5100000;
constexpr long long INF = 1LL << 60;
constexpr int inf = 1000000007;
constexpr long long mod = 1000000007LL;
//constexpr long long mod = 998244353LL;
const long double PI = acos((long double)(-1));

using namespace std;
typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;


int bfs(int start, vector<int>& d, vector<vector<int>>& g) {
	d[start] = 0;
	int last = -1;
	queue<int> q;
	q.push(start);
	while (!q.empty()) {
		int cur = q.front();
		q.pop();
		last = cur;
		for (int next : g[cur]) {
			if (d[next] != -1) continue;
			d[next] = d[cur] + 1;
			q.push(next);
		}
	}
	return last;
}
int main()
{

	cin.tie(nullptr);
	ios::sync_with_stdio(false);

	int n, m, s, t; cin >> n >> m >> s >> t; s--; t--;
	vector<int> cs(n), ct(n);
	vector<vector<int>> g(n);
	for (int i = 0; i < m; i++) {
		int u, v; cin >> u >> v;
		u--; v--;
		g[u].emplace_back(v);
		g[v].emplace_back(u);
	}
	vector<int> ds(n, -1);
	vector<int> dt(n, -1);
	bfs(s, ds, g);
	bfs(t, dt, g);
	ll res = 0;
	int len = ds[t];
	for (int i = 0; i < n; i++) {
		if (dt[i] != -1) {
			ct[dt[i]]++;
		}
		if (ds[i] != -1) {
			cs[ds[i]]++;
		}
	}
	for (int i = 0; i < n; i++) {
		int r = len - i - 2;
		if (r < 0) continue;
		res += 1LL * cs[i] * ct[r];
	}
	cout << res << endl;
}